$\lim_{x \to \infty} \left( \frac{x^3}{3x^2 - 4} - \frac{x^2}{3x + 2} \right)$ is equal to
Adoes not exist
B$1/3$
C0
D$2/9$
Answer & Solution
Correct answer: D. $2/9$
Combine the terms first:
$$\frac{x^3}{3x^2-4}-\frac{x^2}{3x+2}=\frac{x^3(3x+2)-x^2(3x^2-4)}{(3x^2-4)(3x+2)}.$$
Simplifying the numerator,
$$x^3(3x+2)-x^2(3x^2-4)=3x^4+2x^3-3x^4+4x^2=2x^3+4x^2.$$
So the expression becomes
$$\frac{2x^3+4x^2}{(3x^2-4)(3x+2)}.$$
Factor the numerator:
$$\frac{2x^2(x+2)}{(3x^2-4)(3x+2)}.$$
Now divide numerator and denominator by $x^3$:
$$\frac{2\left(1+\frac{2}{x}\right)}{\left(3-\frac{4}{x^2}\right)\left(3+\frac{2}{x}\right)}.$$
Taking $x\to\infty$,
$$\lim_{x\to\infty}\frac{2\left(1+\frac{2}{x}\right)}{\left(3-\frac{4}{x^2}\right)\left(3+\frac{2}{x}\right)}=\frac{2}{3\cdot 3}=\frac{2}{9}.$$
This matches option $\text{D}$.
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